Question 155: Centre of Mass and Rotational Motion

Question

As shown in figure, when a spherical cavity
(centered at O) of radius 1 is cut out of a uniform
sphere of radius R (centered at C), the centre of
mass of remaining (shaded) part of sphere is at
G, i.e, on the surface of the cavity. R can be
detemined by the equation :

Accepted Answer

By concept of COM
m
0
R
0
= m
cavity
R
cavity
$$ \Rightarrow $$ $$\left( {{4 \over 3}\pi {R^3}ho  - {4 \over 3}\pi {{\left( 1 ight)}^3}ho } ight)$$ (2 - R) = $${{4 \over 3}\pi {{\left( 1 ight)}^3}ho }$$ $$ 	imes $$ (R - 1)
$$ \Rightarrow $$ (R
3
Ã¢ÂÂ 1) (2 Ã¢ÂÂ R) = R Ã¢ÂÂ 1
$$ \Rightarrow $$(R
2
+ R + 1) (2 Ã¢ÂÂ R) = 1

Answer

(R
2
+ R – 1) (2 – R) = 1

Answer

(R
2
– R – 1) (2 – R) = 1

Answer

(R
2
– R + 1) (2 – R) = 1

Answer

(R
2
+ R + 1) (2 – R) = 1

Centre of Mass and Rotational Motion

As shown in figure, when a spherical cavity (centered at O) of radius 1 is cut out of a uniform sphere of radius R (centered at C), the centre of mass of remaining (shaded) part of sphere is at G, i.e, on the surface of the cavity. R can be detemined by the equation :

Choose the correct answer: